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bonufazy [111]
3 years ago
10

Which ion is the only negative ion present in an aqueous solution of an Arrhenius base?

Chemistry
1 answer:
Norma-Jean [14]3 years ago
3 0
<span>The correct answer is option 4. It is the hydroxide ions that is produced by an Arrhenius base in water. According to Arrhenius, acid-base reactions are characterized by acids which forms hydronium ions in water and bases which dissociates in aqueous solution to form hydroxide ions. </span>
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When the pressure that a gas exerts on a sealed container changes from 22.5 psi to ? psi, the temperature changes from 110 degre
natta225 [31]
Using Gay-Lussac's Law, pressure is proportional to (absolute) temperature in Kelvin. We first convert the temperature values to Kelvin: 110 C = 383.15 K, while 65 C = 338.15 K.
P1/T1 = P2/T2
22.5/383.15 = P2/338.15
P2 = 19.9 psi
8 0
3 years ago
Heat is added to a chemical system. Which energy will increase?
dmitriy555 [2]

Answer:

the kinetic energy of the reactant molecules(C)

Explanation: i hope this is the correct answer if this is wrong or incorrect please let me know

8 0
2 years ago
Read 2 more answers
If you dissolve 50.0 grams of potassium bromide in 100.0g of water what is the mass percent of the resultant solution?
tatiyna
Mass percent= grams solute/ grams of solution x 100

Mass Percent= (50/ 150)x100= 33.3%
7 0
2 years ago
what is the pressure, in atmospheres, of 2.97 mol h2 gas if it has a volume of 73 liters when the temperature is 298 k? 0.50 atm
Triss [41]
P x V = n x R x T

P x 73 = 2.97 x 0.082 x 298

P x 73 = 72.57492

P = 72.57492 / 73

P = 1.0 atm

hope this helps!




6 0
3 years ago
A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produc
Paul [167]

Answer : The volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of O_2 gas = (740-22.4) torr = 717.6 torr

P_2 = final pressure of O_2 gas at STP= 760 torr

V_1 = initial volume of O_2 gas = 280 mL

V_2 = final volume of O_2 gas at STP = ?

T_1 = initial temperature of O_2 gas = 25^oC=273+25=298K

T_2 = final temperature of O_2 gas = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{717.6torr\times 280mL}{298K}=\frac{760torr\times V_2}{273K}

V_2=242.2mL=0.2422L

Therefore, the volume of O_2(g) produced at standard conditions of temperature and pressure is 0.2422 L

5 0
3 years ago
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